The volume of an $n$-dimensional unit hypercube intersected with the plane $\sum_{i=1}^n x_i=0$.

Question

Let $\mathcal{C}$ be the unit cube in $\mathbb{R}^n$ centred at $\mathbf{0}$, for some integer $n>1$. Consider the following $(n-1)$-dimensional subspace:

$$ V=\left\{(x_1,x_2,\dots,x_n) \in \mathbb{R}^n: \sum_{i=1}^n x_i=0\right\}. $$

I want to calculate the volume of the polytope $\text{Vol}(\mathcal{C} \cap V)$, where by volume we mean the $(n-1)$-dimensional volume as measured in the space $V$ (that is, the volume of the $(n-1)$-dimensional polytope projected onto the plane $V$).

So far, I have determined that such a shape must have vertices whose coordinates are either $0$, $-1/2$ or $1/2$ and such that the coordinates sum to zero. However, beyond this I am uncertain how to calculate the volume of such a shape, and was wondering if anyone had any idea of how to go about solving this problem, or whether it is possible to determine the exact volume of this shape at all for general $n$.

Answer 1

To simplify derivation, we will translate

• unit hypercube $\mathcal{C}$ to $\mathcal{C}' = [0,1]^n$.
• plane $V$ to $V' = \left\{ (x_1,\ldots,x_n) \in \mathbb{R}^n : \sum\limits_{i=1}^n x_i = \frac{n}{2} \right\}$.

Before we start, let's adopt some conventions.

• For any $p \in \mathbb{R}^n$, we will use $p_1,\ldots,p_n$ to denote its components. ie. $p = (p_1,p_2,\ldots,p_n)$.
• For any $p, q \in \mathbb{R}^n$, we will use $p \ge q$ as a shorthand for the relation, $p_i \ge q_i$ for all $i = 1,\ldots, n$.
• For any $p\in \mathbb{R}^n$, let $C_p$ be the cone $\{ x \in \mathbb{R}^n : x \ge p \}$.
• Let $\mathcal{V}$ be vertices of $\mathcal{C}'$ and for $v \in \mathcal{V}$, we will use $(-1)^v$ be a shorthand of $(-1)^{\sum_{i=1}^n v_i}$.
• For any $U \subset \mathbb{R}^n$, we will use $\mathbb{1}_U$ to denote its indicator function.
• For any expression $(\cdots)$, we will use $(\cdots)_{+}$ to denote its positive part, ie. $\max((\cdots),0)$.

Let $m \in \mathbb{R}^n$ be an unit vector with all components positive. For any $t \in \mathbb{R}$, let

• $H_t$ be the half-space $\{ x \in \mathbb{R}^n : m \cdot x \le t \}$.
• $P_t$ be the hyperplane $\{ x \in \mathbb{R}^n : m \cdot x = t \}$.

Let $\mathcal{S}$ be the collection of geometric shapes $\mathbb{R}^n$ that can be formed by finite intersection of open/closed half-spaces. For any $S \in \mathcal{S}$, let $\mu_{n}(S)$ and $\mu_{n-1}(S)$ be the $n$ and $n-1$ dimensional measure of $S$ whenever it make sense.

When $S$ is bounded, aside from finitely many $t$ where $P_t$ contains a facet of closure of $S$, the hyper area of its intersection with $P_t$ is related to the hyper volume of its intersection with $H_t$ by the relation:

$$\mu_{n-1}(S \cap P_t) = \frac{d}{dt}\mu_n(S \cap H_t)$$

Let $\mathcal{C}'' = [0,1)^n$, notice its indicator function is related to those of $C_v$ be the relation.

$$\mathbb{1}_{\mathcal{C}''} = \sum_{v \in \mathcal{V}} (-1)^v \mathbb{1}_{C_v}$$

We find $$\begin{align} \mu_n(\mathcal{C}' \cap H_t) =\mu_n(\mathcal{C}'' \cap H_t) &= \sum_{v \in \mathcal{V}} (-1)^v \mu_n( C_v \cap H_t )\\ &= \sum_{v\in \mathcal{V}} (-1)^v \mu_n( C_0 \cap H_{t-m\cdot v} )\\ &= \frac{1}{n!\prod_{i=1}^n m_i}\sum_{v\in \mathcal{V}} (-1)^v (t - m\cdot v)_{+}^n \end{align}$$ This leads to $$ \mu_{n-1}(\mathcal{C}' \cap P_t) = \frac{d}{dt} \mu_n(\mathcal{C}' \cap H_t) = \frac{1}{(n-1)!\prod_{i=1}^n m_i}\sum_{v\in \mathcal{V}}(-1)^k (t - m\cdot v)_{+}^{n-1} $$ For the problem at hand, we can take all $m_i = \frac1{\sqrt{n}}$ and $P_{\frac{\sqrt{n}}{2}}$ coincides with hyperplane $V'$. This means,

$${\rm Vol}(\mathcal{C} \cap V) = {\rm Vol}(\mathcal{C}' \cap V') = \mu_{n-1}(\mathcal{C}' \cap P_{\frac{\sqrt{n}}{2}} )$$

Notice $m\cdot v = \frac{1}{\sqrt{n}}\sum\limits_{i=1}^n v_i$ and for $k = 0,\ldots, n$, there are $\binom{n}{k}$ vertices with $\sum\limits_{i=1}^n v_i = k$. At the end, we find:

$$\begin{align} {\rm Vol}(\mathcal{C},V) &= \frac{(\sqrt{n})^n}{(n-1)!}\sum_{k=0}^n (-1)^k \binom{n}{k}\left(\frac{\frac{n}{2} - k}{\sqrt{n}}\right)_{+}^{n-1}\\ &= \frac{\sqrt{n}}{(n-1)!}\sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor} (-1)^k \binom{n}{k}\left(\frac{n}{2} - k\right)^{n-1}\tag{*1} \end{align} $$

As a double check, let's look at the special case $n = 3$.

$V$ will be a regular hexagon will side $\frac{1}{\sqrt{2}}$. So its area will be $\frac{3\sqrt{3}}{2}\left(\frac{1}{\sqrt{2}}\right)^2 = \frac{3\sqrt{3}}{4}$.

For comparison, the formula $(*1)$ reproduces $\frac{\sqrt{3}}{2!}\left(\left(\frac32\right)^2 - 3\left(\frac12\right)^2\right) = \frac{3\sqrt{3}}{4}$ as expected.

Answer 2

This can be written as a graph $x_n= -\sum_1^{n-1} x_i$ over a subset $D$ of the corresponding $n-1$ dimensional hypercube, $D$ to be determined. If you happen to know that the area of the graph of a function $f: D\rightarrow \mathbb{R}$ can be calculated as $$A(f, D) = \int_D \sqrt{1+ ||\nabla f||^2}d\bar x $$ (with the notation $\bar x=(x_1,\ldots, x_{n-1})\in D \subset \mathbb{R}^{n-1}$) you should be able to calculate the area of the surface in question once you figure out to which subset of the $n-1$ dimensional cube the hyperplane projects.